Haloalkanes Methods of Preparation

Haloalkanes Methods of Preparation

• Nature of C-X Bond: Since halogen atoms are more electronegative than carbon, the carbon halogen bond of alkyl halide is polarised; the carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge.

• Haloalkanes Methods of Preparation: Alkyl halides are best prepared from alcohols, which are easily accessible. The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides, or thionyl chloride.
• From alcohols:

 a). By the action of concentrated halogen acids on alcohol in presence of anhydrous ZnCl₂ as a catalyst.
R − OH + HX → R − X + H₂O
Rate: HI > HBr > HCl
Tertiary (3°) > Secondary (2°) > Primary (1°)

• Since HCl is least reactive it requires the presence of anhydrous ZnCl₂ for the reaction
• This reaction forms the basis of the Lucas Test which is used to distinguish between 1⁰, 2^0, and 3⁰ alcohols.

• Reactions of primary and secondary alcohols with HI require the presence of an. ZnCl₂, while tertiary alcohols do not require the catalyst.

b). By the action of phosphorus halides with alcohols.
R − OH + PX₅ → R − X + POX₃ + HX
3 R − OH + PX₃ → 3R − X + H₃PO₃

• Chloro alkanes are obtained by the direct reaction with PCl₅ or PCl₃.
• PI₃ and PBr₃ are generated in situ by the reaction of red phosphorus with iodine or bromine

c). Alkyl chlorides are obtained by the action of thionyl chloride (SOCl₂) with alcohol.
R − OH + SOCl₂ → R − Cl + SO₂ (g) + HCl(g)
Among these methods, the reaction with thionyl chloride (SOCl₂) is preferred, since the byproducts are gases and are easily escaped from the reaction medium.
Among the Haloalkanes and Haloarenes, the preparation of aryl halide from phenols becomes difficult since the carbon-oxygen bond in phenols has a partial double bond character and hence is difficult to break.

• From Hydrocarbons:

(a) By free radical halogenation: Free radical chlorination or bromination of alkanes in presence of light gives a complex mixture of isomeric mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield of any one compound is low.
CH₄ + Cl₂ → CH₃Cl + CH₂Cl₂ + CHCl₃ + CCl₄

• Order of ease of replacement of a hydrogen atom is:

Tertiary > Secondary > Primary > CH₄

• Iodination is also difficult because of the low reactivity of iodine. Secondly, HI which is formed during the reaction is a strong reducing agent and reduces alkyl iodide back to hydrocarbon. Thus, iodination can only be carried out in the presence of oxidizing agents like HNO₃, HIO_3, or HIO₄. These oxidizing agents react with HI and oxidize it to iodine thus preventing the reverse reaction.

CH₃CH₃ + I₂ → CH₃CH₂I + HI
5HI + HIO₃ → 3I₂ + 3H₂O

• From alkenes:
• a). Addition of hydrogen halides: An alkene is converted to a corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide, or hydrogen iodide.

CH₂=CH₂ + HBr  → CH₃ — CH₂Br + HBr
In the case of unsymmetrical alkenes, the addition takes place according to Markownikoff’s rule. [The rule states that “when an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the addendum (adding molecule) gets attached to the carbon-containing a lesser number of hydrogen atoms”)
CH₃-CH=CH₂ + HBr  →  CH₃-CH₂-CH₂Br + CH₃-CHBr-CH₃
                                                    (minor)                 (major)

b).  Addition of Halogen:  Alkenes add halogen to form vicinal dihalides (2 halogen atoms on adjacent C atoms). For e.g. addition of bromine in CCl₄ to an alkene results in the formation of vicinal dibromides and also in the discharge of the reddish-brown colour of Br₂ in CCl₄. So this is used as a test for unsaturation.                                                       
                CH₂=CH₂ + Br₂  →  CH₂Br – CH₂Br
                                                (1,2-dibromoethane)

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